Solution
= Solution
Using the exponential definition of <hyperbolic tangent>, set $q=e^{2z}$:
$$
\frac{q-1}{q+1}=-i,\qquad(1+i)q=1-i,\qquad q=-i.
$$
Here $q+1=1-i\ne0$, so the resulting points are not <poles>. Taking all values of the <complex logarithm> gives $2z=i(-\pi/2+2\pi k)$. Thus
$$
\boxed{z=-\frac{i\pi}{4}+i\pi k,\qquad k\in\mathbb Z.}
$$
Equivalently $\tanh(iu)=i\tan u$, so these are exactly the imaginary-axis solutions of $\tan u=-1$.